Hess's Law

Dodi1991

Hey there, been having some trouble with getting this thing sorted. The concept seems so simple ... but I still can't get my head around it. Can someone please show me an example of a Hess's law with Heat of Combustion and an example of Heat of Formation please?

Perhaps with a couple of explanations too with each steps, I'll be grateful :smile:

Reply 1

Andylol

14

Hess's law states that the enthalpy change for a chemical reaction is the same, whatever route is taken from reactants to products.

Reply 2

Dodi1991

OP

I understand what it means...I'd just like to see a couple of examples to make it a little clearer.

Reply 3

CJN

17

C + 2H2 = CH4 is the overall reaction

But it wouldn't work in reality you would get many hydrocarbons not just methane.

So it would be:

Combustion: C -> CO2
Combusion: 2H2 -> 2H2O

Then CO2 + 2H2O -> CH4

The first reaction is only one step but isn't possible. So in reality people would use the second method consisting of three different steps to do this, but overall they both use the same amount of energy/the enthalpy change is the same overall.

Most of the Hess' law cycles/calculations work on the same idea: that it is hypothetically possible to go from both the reactants and the products to the same intermediate stage.

Hess 1: Via the elements

This is using the enthalpy of formation. Breaking the reactants into their constituent elements (at standard state) is the reverse of the enthapy of formation of the reactants.
Making the products from the (newly formed) elements at standard state is the enthapy of formation of the products.

Thus: enthapy of reaction = formation enthalpy products - formation enthapy reactants.

Hess 2: via the individual atoms

Breaking the reactants into the constituent atoms is the bond enthalpy of all of the constituent bonds (endothermic).
Making the products from the constituent atoms is the reverse of the bond enthalpy (exothermic)

Thus: enthalpy of reaction = bond enthalpies of reactants - bond enthalpies of products

Hess 3: via combustion products

The reactants burn to make the combustion products (usually CO2 and H2O), an exothermic (delta H negative) reaction.
The route from combustion products to reaction products is the reverse of combustion of the reaction products (i.e. endothermic)

Thus reaction enthalpy = enthapy of combustion of the reactants - reaction enthalpy of the products.

Reply 5

prettyboy_bn

u should have some quns and bring on here , we will try to draw a diagram and solve it :biggrin:

Reply 6

Dodi1991

OP

Here is one example of a Hess's law question I'd like to know how to do :smile:

Enthalpy changes of combustion can be used to determine enthalpy changes of formation.

Combustion of Butane:

C4H10 + 6\frac{1}{2}O2 ---> 4CO2 + 5H20

i) Write the equation for the standard enthalpy change of formation of butane, C4H10. Include state symbols in your answer.

ii) Use the following data to calculate the standard enthalpy change of formation of butane.

standard enthalpy change of combustion KJ mol^-1
Carbon -394
Hydrogen -286
Butane -2877

Dodi1991

Here is one example of a Hess's law question I'd like to know how to do :smile:

Enthalpy changes of combustion can be used to determine enthalpy changes of formation.

Combustion of Butane:

C4H10 + 6\frac{1}{2}O2 ---> 4CO2 + 5H20

i) Write the equation for the standard enthalpy change of formation of butane, C4H10. Include state symbols in your answer.

ii) Use the following data to calculate the standard enthalpy change of formation of butane.

standard enthalpy change of combustion KJ mol^-1
Carbon -394
Hydrogen -286
Butane -2877

consider the combustion produdcts to be an intermediate step in the reverse of the formation enthalpy equation for butane:

C4H10 + 61/2O2 ---> deltaH combustion -- 4CO2 + 5H2O <--- delta H combustion <-------- 4C + 5H2

you should be able to see that to go backwards from 4C + 5H2 = (combustion enthalpies of 4C + combustion enthalpy of 5H2) - Combustion enthalpy of C4H10

Reply 8

saminasherulxx

Original post by charco

consider the combustion produdcts to be an intermediate step in the reverse of the formation enthalpy equation for butane:

C4H10 + 61/2O2 ---> deltaH combustion -- 4CO2 + 5H2O <--- delta H combustion <-------- 4C + 5H2

you should be able to see that to go backwards from 4C + 5H2 = (combustion enthalpies of 4C + combustion enthalpy of 5H2) - Combustion enthalpy of C4H10